# 532. K-diff Pairs in an Array 数组中的 k-diff 数对

@TOC

## # 题目描述

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:

``````Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.
``````

Example 2:

``````Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
``````

Example 3:

``````Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
``````

Note:

1. The pairs (i, j) and (j, i) count as the same pair.
2. The length of the array won't exceed 10,000.
3. All the integers in the given input belong to the range: [-1e7, 1e7].

## # 解题方法

### # 字典

``````import collections
class Solution(object):
def findPairs(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
counter = collections.Counter(nums)
for num in set(nums):
if k > 0 and num + k in counter:
if k == 0 and counter[num] > 1:
``````

``````class Solution(object):
def findPairs(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
res = 0
if k < 0: return 0
elif k == 0:
count = collections.Counter(nums)
for n, v in count.items():
if v >= 2:
res += 1
return res
else:
nums = set(nums)
for num in nums:
if num + k in nums:
res += 1
return res
``````

C++版本的代码如下：

``````class Solution {
public:
int findPairs(vector<int>& nums, int k) {
unordered_map<int, int> m;
for (int num : nums) {
m[num]++;
}
int res = 0;
for (const auto &it : m) {
if (k == 0 && it.second >= 2) {
res ++;
} else if (k > 0 && m.count(it.first + k)) {
res ++;
}
}
return res;
}
};
``````

## # 日期

2018 年 2 月 4 日 2018 年 11 月 27 日 —— 最近的雾霾太可怕